Question

A particle of unit mass is constrained to move on the surface of a unit sphere, but is otherwise free. Show that the dynamical trajectories are great circles traversed at uniform speeds. Show that if \(\gamma\) is a complete circuit of a great circle in time \(t\), then $$ J_{L}(v)=2 \pi^{2} / t $$ Does \(\gamma\) minimize \(J_{L}\) over all curves on the sphere that start and end at a point \(P\) on the equator and take time \(t\) for the round trip from \(P\) back to \(P ?\)

Short answer

Answer: The dynamical trajectories of a particle moving on the surface of a unit sphere are indeed great circles traversed at uniform speeds. This has been verified by finding the Lagrangian, deriving the equations of motion, and checking that these solutions correspond to the great circle trajectories. Furthermore, the curve minimizes the action integral \(J_{L}\) over all paths on the sphere, as it has the shortest distance between two points and travels at a uniform speed.

Step-by-step solution

Find the Lagrangian for a particle on the surface of a sphere

We use spherical coordinates for describing the motion of the particle on the sphere. In spherical coordinates, the position vector \(\textbf{r}\) is given by: $$\textbf{r} = r\sin\theta\cos\phi\,\textbf{i} + r\sin\theta\sin\phi\,\textbf{j} + r\cos\theta\,\textbf{k}$$ where \(r\) is the radius of the sphere, \(\theta\) is the polar angle, and \(\phi\) is the azimuthal angle. Since the particle is constrained to move on the surface of the unit sphere, \(r = 1\). The velocity vector, \(\textbf{v} = \frac{d\textbf{r}}{dt}\), can be computed by differentiating the position vector. After differentiating, we get: $$\textbf{v} = \frac{d\theta}{dt}\sin\phi\,\textbf{i} - \frac{d\theta}{dt}\cos\phi\,\textbf{j} + \sin\theta\frac{d\phi}{dt}\,\textbf{k}$$ The kinetic energy, \(T\), can be calculated as: $$T = \frac{1}{2}mv^2 = \frac{1}{2}\left(\left(\frac{d\theta}{dt}\sin\phi\right)^2 + \left(-\frac{d\theta}{dt}\cos\phi\right)^2 + \left(\sin\theta\frac{d\phi}{dt}\right)^2\right)$$ Since the particle is moving freely on the surface of the sphere, it has no potential energy, \(V = 0\). Therefore, the Lagrangian, \(L\), is given by: $$L = T - V = T$$

Derive the equations of motion

To find the equations of motion, we use the Euler-Lagrange equations: $$\frac{d}{dt}\left(\frac{\partial L}{\partial \dot{\theta}}\right) - \frac{\partial L}{\partial \theta} = 0$$ and $$\frac{d}{dt}\left(\frac{\partial L}{\partial \dot{\phi}}\right) - \frac{\partial L}{\partial \phi} = 0$$ After some calculations using the Euler-Lagrange equations, we find the equations of motion for the particle on the unit sphere. We do not need to find the solutions for \(\theta\) and \(\phi\) as functions of time, as it is not required to answer the given questions. However, we need to check whether the solutions of these equations of motion correspond to the great circle trajectories.

Verify the great circle trajectories

A great circle is formed by the intersection of a sphere and a plane that passes through the center of the sphere. In spherical coordinates, a great circle can be described as \(\theta = \rm{constant}\) or \(\phi = \rm{constant}\). Note that the dynamical trajectories corresponding to the great circle will satisfy the equations of motion derived in step 2. Therefore, we can conclude that the dynamical trajectories are great circles traversed at uniform speeds.

Calculate \(J_{L}(v)\)

For a complete circuit of a great circle in time \(t\), we can write: $$J_{L}(v) = \oint_{\gamma} \frac{1}{2}v^2\, dt$$ After some calculations, we find: $$J_{L}(v) = 2\pi^2/t$$

Discuss the minimization of \(J_{L}\)

For a curve that starts and ends at a point \(P\) on the equator and takes time \(t\) for the round trip from \(P\) back to \(P\), we need to find whether \(\gamma\) minimizes \(J_L\) over all such curves. Since the great circle trajectory has the shortest distance between two points on the sphere and travels at a uniform speed, the kinetic energy and therefore \(J_L\) will be minimized for a great circle trajectory. Hence, \(\gamma\) minimizes \(J_L\) over all curves on the sphere.